joint pmf

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November 20th 2010, 01:09 PM
chutiya

joint pmf

if X and Y are discrete RVs with pmf $p(x,y) = a\; \dfrac{k!}{x!y!(k-x-y)!}$ where x and y are $non \; negative \; integers \; and\; x+y\leq k$ .

I want to find the value of a.

I know that I need to show that the pmf sums up to 1.

$\sum_x \sum_y p(x,y)=1$
since x and y are non negatve: 0<=x+y<=k
here i am getting confused in where x and y range from.
can anyone help me how to solve this?
November 21st 2010, 04:42 AM
Robb

$\sum_{x=0}^{k} \sum_{y=0}^{k-x} p(x,y)=1$

so,
$a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!}{x!y!(k-x-y)!} = a \cdot \sum_{x=0}^{k} \dfrac{k!\cdot 2^{k-x}}{x!(k-x)!}=a\cdot 3^k =1$

Hence $a=\left(\frac{1}{3}\right)^k$
November 21st 2010, 06:26 AM
chutiya

Thank you so much.

But how do we get $\sum_{y=o}^{k-x}\dfrac{1}{y!(k-x-y)!}= \dfrac{2^{k-x}}{(k-x)!}$

is this a gamma function? Could you explain please.
November 22nd 2010, 04:54 AM
Robb

Refer to the section on series of binomial coefficients
Binomial coefficient - Wikipedia, the free encyclopedia

$a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!}{x!y!(k-x-y)!} = a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!\cdot \left(k-x\right)!}{x!y!(k-x-y)!(k-x)!} =a \cdot \sum_{x=0}^{k} \dfrac{k!}{x!(k-x)!}\sum_{y=0}^{k-x} \dfrac{\left(k-x\right)!}{y!(k-x-y)!}$

and additionally from binomial theorem
$(Y+X)^n = \sum_{k=0}^n {n \choose k} X^{n-k}Y^K$

So using this formula, we can write
$a \cdot \sum_{x=0}^{k} \dfrac{k!\cdot 2^{k-x}}{x!(k-x)!}= a \cdot \sum_{x=0}^\infty {k \choose x} 2^{k-x}1^k=a \cdot \left(2+1\right)^k$